[leet code] Populating Next Right Pointers in Each Node

Given a binary tree

struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to

NULL

.

Initially, all next pointers are set to

NULL

.

Note:

You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

Given the following perfect binary tree,

1
/  \
2    3
/ \  / \
4  5  6  7

After calling your function, the tree should look like:

1 -> NULL
/  \
2 -> 3 -> NULL
/ \  / \
4->5->6->7 -> NULL

思路总结:

看到树结构, 第一反应要用递归

然后开始罗列情况:

1. root node直接指向null

2. 其他情况在递归中实现

2.1 如果该节点有左子节点(根据题目定义同时也会有右子节点)

2.1.1 左子节点指向右子节点

2.1.2 右子节点指向节点.next的左直接点(题目example中的节点5->6); 如果右子节点已经是该层最右节点(通过节点.next == null判断), 则该右子节点指向null

2.2 如果该节点没有子节点则什么都不做, 返回上一层递归

remark: 写code的时候另外新建了helper function为了把第一种情况从递归中分离出来, 方便理解.

/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
*     int val;
*     TreeLinkNode left, right, next;
*     TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
if (root != null){
// root.next always -> null
root.next = null;
helper(root);
}
}

public void helper(TreeLinkNode node){
if (node.left != null){
//left child -> right child
node.left.next = node.right;
//right child -> null if the right most
if(node.next == null)
node.right.next = null;
//right child -> right node's left child
else
node.right.next = node.next.left;
helper(node.left);
helper(node.right);
}
}
}